Let $x,$ $y,$ and $z$ be nonnegative real numbers such that $x + y + z = 2.$  Find the maximum value of
\[(x^2 - xy + y^2)(x^2 - xz + z^2)(y^2 - yz + z^2).\]
Without loss of generality, we can assume that $z \le x$ and $z \le y.$  Then
\[(x^2 - xy + y^2)(x^2 - xz + z^2)(y^2 - yz + z^2) \le (x^2 - xy + y^2) x^2 y^2.\]By AM-GM,
\begin{align*}
x^2 y^2 (x^2 - xy + y^2) &= \frac{4}{9} \left( \frac{3}{2} xy \right) \left( \frac{3}{2} xy \right) (x^2 - xy + y^2) \\
&\le \frac{4}{9} \left( \frac{\frac{3}{2} xy + \frac{3}{2} xy + (x^2 - xy + y^2)}{3} \right)^3 \\
&= \frac{4}{9} \left( \frac{x^2 + 2xy + y^2}{3} \right)^3 \\
&= \frac{4}{9} \cdot \frac{(x + y)^6}{27} \\
&\le \frac{4}{243} (x + y + z)^6 \\
&= \frac{256}{243}.
\end{align*}Equality occurs when $x = \frac{4}{3},$ $y = \frac{2}{3},$ and $z = 0,$ so the maximum value is $\boxed{\frac{256}{243}}.$